Finding Volumes: Washer/Disc Methods
Date: 01/05/98 at 22:21:22 From: Leeanne Nagle Subject: Calculus - washer/disc methods Dear Dr. Math, I am a current senior in my high school in Massachusetts. I find my Calculus course this year to be extremely difficult and not well taught. At the moment I am having difficulties in the washer and disc methods of finding volumes. Such is the case for the problem y = x^2, x = 3 about the y axis. I don't know where to begin. I also find it difficult to determine the surface area in a problem such as y = ((x^(3/2))/3)- x^(1/2) o<_x<_3. If there is any way you can help I would greatly appreciate it. Leeanne Nagle
Date: 01/07/98 at 11:39:04 From: Doctor Jaffee Subject: Re: Calculus - washer/disc methods Hi Leeanne, I found my Calculus course extremely difficult, too, and I particularly struggled with the washer and disk methods for finding volumes, but I have a few ideas that I hope will make it a little easier for you. First of all, making a good 3-dimensional drawing of the problem is far and away the best start. Now that's going to be tough over the computer, but let me describe the drawing and you can actually construct the picture. Imagine you are standing in a huge empty room (sort of like the holodeck on Star Trek before they run a program). Now across the middle of the floor is a thin line and at the midpoint of the line is a thin vertical pole going all the way to the ceiling. The line is going to be the x-axis and the pole is the y-axis. Now, you need to stand away from the line and ask the holodeck computer to generate the line x = 3. You will see another thin vertical pole go through the line on the floor 3 units from the origin. Tell the holodeck computer to generate the parabola y = x^2 and you will see a parabola rise up from the floor, its vertex at the origin, and intersecting the line x = 3 at the point (3,9). Can you draw all of this? If so, we can continue. Now, the x-axis, the parabola, and the line x = 3, intersect to form a figure much like a right triangle except the hypotenuse is not a segment, it is a curve that is part of the parabola. Pick a random point on the line x = 3 somewhere between the x-axis and the point where the line and parabola cross. Let's call this point P. Now revolve P around the y-axis. The result should be a circle parallel to the floor. The center of the circle is on the y-axis at a point which is the same height as P. The radius of the circle is 3. Can you see this circle in your drawing? If so, we can move on to the next step. Now find the point on the parabola that is the same height as point P. We'll call this point Q. Revolve Q around the y-axis and the result should be a circle inside the first circle. It, too, is parallel to the ground and is centered on the y-axis. Its radius, however, depends on how high it is. In any case, the radius of this circle is equal to the x-number at Q. You should be able to see that if you take the area of the big circle (PI*9) and subtract the area of the little circle (PI*x^2), the result will be the area of the "washer" between the parabola and the line x = 3. Now if you give this flat washer a little thickness, it will become a disk. Let's call the thickness Delta y, which I will abbreviate as Dy. (In your textbook the delta looks like a triangle). So, the volume of this disk is the area of the base of the disk times its height, or (PI*9 - PI*x^2)*Dy. If you were to start all over again and pick a different location for point P, and go through the steps to make another disk, and keep on doing that until you had nearly filled the region with these disks, you could add up their volumes and have an approximation for the volume of revolution. Of course, if the disks were made with less thickness, and you had more of them, you could get an even better approximation. The sum of the volumes of all these disks is called a Riemann Sum. Finally, if you were to calculate the sum of the infinitely many disks required when the limit of the thickness of the disks approached zero, you would be finding the integral of the function. The Dy would become dy and the x^2 would be renamed y, and since the parabola and the line x = 3 cross at (3,9) the limits of integration would be 0 and 9. So, the volume would be the integral from 0 to 9 of (PI*9 - PI*y)dy. I think you can figure it out from there. You should end up with (81/2)*PI for your final answer. I realize we're really doing this the hard way, but if you have a few classmates with you as you go through my explanation, I think that you can come up with a good drawing and find the solution. Now, in the second problem I'm assuming that you want to revolve the curve around the x-axis from 0 to 3. If you have a cone and make two horizontal slices, the surface between the slices is called a frustum. The surface area of a frustum is 2*PI*r*l, where r is the average of the radii of the two circles where the slices were made, and l is the slant length of the frustum. Finding the surface area for your problem is similar. If you were to make a series of vertical slices form 0 to 3, the result would be a lot of surfaces, each of which was approximately a frustum whose radius was the y number on the curve, and whose slant length was the arc length of the curve between the two slices. If you were to add up all the areas you would get a good approximation for the surface area of revolution. If you were to make the slices a lot closer, however, you would get an even better approximation. And finally, if you were to make the limit of the distance between the slices equal to zero, requiring infinitely many slices, then you would have the exact area. But that's what the integral is. So, take the integral from 0 to 3 of 2*PI*(((x^(3/2)/3) - x^(1/2))*SQRT(1 - (f'(x))^2)dx and you will have your answer. There is some tricky algebra involved in solving this, but you should be able to get some help from your teacher and classmates on that. At any rate, I hope I've provided you with the means to have a better understanding on what's going on in the problem. Good luck in your studies. -Doctor Jaffee, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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